Question

find the sq. root of 5-12i

Answer 1

let me know if u dont get any step

Answer 2

$let \sqrt{ - 5 + 12i} = a + ib$

$- 5 + 12i = (a + ib {)}^{2} = {a}^{2} - {b}^{2} + 2iab$

$We \: will \: get \: {a}^{2} - {b}^{2} = - 5, \: 2ab = 12$

${a}^{2} - \frac{36}{ {a}^{2} } = - 5.......b = \frac{6}{a}$

${a}^{4} + 5 {a}^{2} - 36 = 0$

$( {a}^{2} - 4)( {a}^{2} + 9) = 0$

$a = ±2 \: and \: ±3$

$we \: have \: a = 2 \: and \: b = 3$

$a = - 2 \: and \: b = - 3$

$b = \frac{6}{a} = ±3 so \sqrt{ - 5 + 12i}$

$= ±(2 + 3i)$

hope it's helps you ❤️